HUST 1599 - Multiple(动态规划)

1599 - Multiple

时间限制：2秒 内存限制：64兆

461 次提交 111 次通过

题目描述

Rocket323 loves math very much. One day, Rocket323 got a number string. He could choose some consecutive digits from the string to form a number. Rocket323 loves 64 very much, so he wanted to know how many ways can he choose from the string so the number he got multiples of 64 ?

A number cannot have leading zeros. For example, 0, 64, 6464, 128 are numbers which multiple of 64 , but 12, 064, 00, 1234 are not.

输入

Multiple cases, in each test cases there is only one line consist a number string.

Length of the string is less than 3 * 10^5 .

Huge Input , scanf is recommended.

输出

Print the number of ways Rocket323 can choose some consecutive digits to form a number which multiples of 64.

样例输入

64064

样例输出

5

提示

There are five substrings which multiples of 64.

[64]064

640[64]

64[0]64

[64064]

[640]64

来源

Problem Setter : Yang Xiao

思路：

根据同余定理，每次枚举到一个数，都把前面的存在的余数乘以10加上这个数再对64取余，就产生新的余数，最后统计余数是0的个数有多少

#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>

using namespace std;
char a[300005];
long long int dp[2][65];
long long int tag[65];
long long int ans;
long long int res;
int now;
int main()
{
    while(scanf("%s",a)!=EOF)
    {
        ans=0;res=0;
        int len=strlen(a);
        memset(dp,0,sizeof(dp));
        now=0;
        for(int i=0;i<len;i++)
        {
            for(int j=63;j>=0;j--)
                dp[now^1][j]=0;
            for(int j=63;j>=0;j--)
            {
                int num=(j*10+a[i]-'0'+64)%64;
                dp[now^1][num]+=dp[now][j];

            }
            if(a[i]!='0')
                dp[now^1][a[i]-'0']++;
            else
                ans++;
            res+=dp[now^1][0];
            now^=1;

        }
        printf("%lld\n",res+ans);
    }
    return 0;
}